The square root of $x$ is greater than 2 and less than 4. How many integer values of $x$ satisfy this condition?
Answer: We have: $4 > \sqrt{x} > 2$. Squaring, we get $16 > x > 4$.  Thus, the integers from 15 to 5, inclusive, satisfy this inequality. That's a total of $15-5+1=\boxed{11}$ integers.